mathematics-physics-wiki/docs/en/mathematics/ordinary-differential-equations/first-order-ode.md

# First-order differential equations

## First-order linear differential equations

A first-order **linear** differential equation is one of the type

$$
\frac{dy}{dx} + p(x) y = q(x).
$$

where $p(x)$ and $q(x)$ are given functions, which may be assumed to be continuous. The equation is called **nonhomogeneous** unless $q(x)$ is dentically zero. The corresponding **homogeneous** equation

$$
\frac{dy}{dx} + p(x)y = 0,
$$

is separable and so is easily solved. By seperation of variables

$$
\frac{dy}{dx} = -p(x)y \implies \int \frac{1}{y}dy = \int p(x)dx.
$$

Though, pay attention to absolute values.

There are two methods for solving nonhomogeneous equations.

### Integration factor

The first method is by using an integrating factor. Let $\mu(x)$ be an antiderivative of $p(x)$ and multiply the equation by $e^{\mu(x)}$.

$$
\begin{array}{ll}
e^{\mu(x)} \frac{dy}{dx} + e^{\mu(x)} p(x) y = e^{\mu(x)} q(x) &\implies \frac{d}{dx}(e^{\mu(x)} y) = q(x) e^{\mu(x)}, \\
&\implies e^{\mu(x)} y = \int q(x) e^{\mu(x)}dx, \\
&\implies y(x) = e^{-\mu(x)} \int q(x) e^{\mu(x)}dx.
\end{array}
$$

### Variation of the constant

The second method is by using a variation of a constant. Let $\mu(x)$ be an antiderivative of $p(x)$ and solve the homegeneous equation

$$
\frac{dy}{dx} + p(x)y = 0 \implies y(x) = k e^{-\mu(x)}.
$$

Try:

$$
y'(x) + p(x) y(x) = q(x) = k'(x) e^{\mu(x)},
$$

thus $k'(x) = q(x) e^{\mu(x)}$.

#### Example

Solve $\frac{dy}{dx} + 2xy = x$ with $y(0) = 3$.


First solving the homogeneous equation

$$
\begin{array}{ll}
\frac{dy}{dx} + 2xy = 0 &\implies \int \frac{1}{y} dy = -2\int xdx
\end{array}
$$